What is the integral of 1/(xsqrt(4x^2-16)) ?

The integral of 1/(xsqrt(4x^2-16)) is 1/4 arcsec(x/2)+C

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integral of 1/(xsqrt(4x^2-16))

\int \frac{1}{x 4 x ^{2} - 16} d x

\frac{1}{4} arcsec (\frac{x}{2}) + C

Solution steps

\int \frac{1}{x 4 x ^{2} - 16} d x

Apply Trigonometric Substitution

: \int \frac{1}{4} d u

= \int \frac{1}{4} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{4} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{2}{4} x)

= \frac{1}{4} \cdot 1 \cdot arcsec (\frac{2}{4} x)

Simplify

\frac{1}{4} \cdot 1 \cdot arcsec (\frac{2}{4} x) : \frac{1}{4} arcsec (\frac{x}{2})

= \frac{1}{4} arcsec (\frac{x}{2})

Add a constant to the solution

= \frac{1}{4} arcsec (\frac{x}{2}) + C

y = 7 + 5 x^{2} - 2 x^{3}, (2, 11)

x \to 2 lim (\frac{x + 4}{x - 2})

\frac{d}{d x} (ln (cos (\frac{1}{x})))

\frac{d y}{d x} = e^{5 x} + 8 y

y^{3} - (2 x + 4) + 3 x y^{2} y^{^{'}} = 0

What is the integral of 1/(xsqrt(4x^2-16)) ?
The integral of 1/(xsqrt(4x^2-16)) is 1/4 arcsec(x/2)+C